Question

Measure of angle of elevation of top of towere 57root3high from a point 57 from the foot of tower

Answer 1

60° is the angle 50√3 /50= √3 = tan 60°

Answer 2

Height of the tower ( AB ) = 57√3

Distance from foot of the tower to

observer ( BC ) = 57

In ∆ABC , <B = 90°

Angle of elevation = <ACB

tan <ACB = AB/BC

= ( 57√3 )/57

= √3

= tan 60°

Therefore ,

<ACB = 60°

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